Good demonstration. Thanks too to @maisonville7656 for these interesting remarks. Here another demonstration, a bit different from 48 hours ago before you deleted a similar video. Like you at 8:22, Q is the incenter of the right-angled triangle ABC (Q center of the inscribed circle with radius=h) Then : h=(AB+BC-AC)/2 AB=AC*cos(2*a) and BC=AC*sin(2*a) h=AC/2*(cos(2*a)+sin(2*a)-1) PB=x/2 and BR=108/x then we must understand that : PB*BR=x/2*108/x=54 PB=BC*tan(b)=AC*sin(2*a)*tan(45°-a) BR=AB*tan(a)=AC*cos(2*a)*tan(a) PB*BR=AC^2*sin(2*a)*tan(45°-a)*cos(2*a)*tan(a) We can notice that : sin(2*a)*tan(a)=1-cos(2*a) cos(2*a)*tan(45°-a)=1-sin(2*a) PB*BR=AC^2*(1-cos(2*a))*(1-sin(2*a)) PB*BR=1/2*AC^2*(cos(2*a)+sin(2*a)-1)^2 PB*BR=1/2*(2*h)^2 PB*BR=2*h^2 h=sqrt(PB*BR/2) h=sqrt(54/2) h=3*sqrt(3) A|ACQ|=1/2*AC*h A|ACQ|=1/2*26*3*sqrt(3) A|ACQ|=39*sqrt(3) ************************ Nota Bene : We can demonstrate in other way, geometrically that : PB*BR=AC^2*(1-cos(2*a))*(1-sin(2*a)) Let's introduce 2 points : S orthogonal projection of P on (AC) T orthogonal projection of R on (AC) Then, PBC and PSC are 2 identical right-angled triangle (same angles b, 90°-b, 90° and same hypothenuse PC) Then, ABR and ATR are 2 identical right-angled triangle (same angles a, 90°-a, 90° and same hypothenuse AR) Then : PB=PS ; BR=RT ; SC=BC ; AT=AB TC=AC-AT=AC-AB=AC-AC*cos(2*a)=AC*(1-cos(2*a)) AS=AC-SC=AC-BC=AC-AC*sin(2*a)=AC*(1-sin(2*a)) AS*TC=AC^2*(1-cos(2*a))*(1-sin(2*a)) APS and RTC are 2 right-angled triangle. PS=AS*tan(2*a) and RT=TC*tan(2*b)=TC/tan(2*a) PS*RT=AS*tan(2*a)*TC/tan(2*a) PS*RT=AS*TC PB*BR=AS*TC PB*BR=AC^2*(1-cos(2*a))*(1-sin(2*a))
Q is the center of the in-circle. angle B being 90 degrees, BQ must be equal to sqr(2)*radius of in-circle. Figuring out that triangles BQR and BQP are similar was more work, but can easily be done. BR/BQ=BQ/BP gives 2*radius^2=54. The radium of the in-circle is the height of triangle AQC. Nice problem. Most of the work was to demonstrate BQR and BQP triangles are similar.
Area of right triangle BPR: A = ½ y.z = ½ .x/2 .108/x = 27 m² Now, let's find the square adjacent to previous right triangle, with same area. A = 27 m² = s² --> s =√27m This square has opposite vertices B and Q, and its side is equal to height of blue shaded triangle: h = s = √27 = 3√3 m Area of blue shaded triangle: A = ½b.h = ½ 26 3√3 A = 67,54 m² ( Solved √ ) [ There's a little mistake in video, y=x/2 m and not y=x/(2m) ]
That's weird. You seem to demonstrate the exercise in a few lines, with too few calculations. How do you prove that the square whose side is h, which therefore has [BQ] as diagonal, has the same area as the triangle BPR ?
@@matthieudutriaux Thanks four your comment. For any angle "α" and "β", while the triangle is a right triangle, then always BQ is the diagonal of a square, with sides aligned with legs of triangle. Because always location angle of BQ is 45° respect to right triangle legs. And, of course, side "s" is equal to "h" easy to see, property of bisected triangles. Regarding the area of the minor right triangle PBR, always Is equal to area of the mentioned square with diagonal BQ. You can find the demonstration in video, minute 13:11, where says " y.z= 2.h² ", what is the same I wrote in my comment. This is a property of bisected right triangles, and we already know it in advance. If you have any doubt, just modify the ratio y/z , or what is the same, modify angles "α" and "β". and you will verify that always this rule is fulfilled.
@@marioalb9726 I understand all the video of @MathandEngineering Then, i understand the demonstration at 13:11 when "y*z=2*h^2" (little error : not y*x) I have demonstrated myself (by 2 methods) that PB*BR=y*z=2*h^2. But me and @MathandEngineering and others make the demonstration. I am not agree with you when you say : "This is a property of bisected right triangles, and we already know it in advance." All this exercise is a simple right-angled triangle. Q is the center of bisectors. Area |ACQ|=1/2*base*h=1/2*26*h The difficulty of the exercise is to prove that y*z=2*h^2 I know we can change the ratio y/z and we will always have y*z=2*h^2 (but if we change the ratio y/z and if AC=26 then y*z=2*h^2 can be different from 54) You may be right but you consider this relationship "y*z=2*h^2" as a mathematical theorem of "bisected right triangles" that we learn and know in advance. The goal of the exercise is to demonstrate it. If we already know it, it is true that the answer is faster.
@@matthieudutriaux Why do you say "the goal of this exercise is to demonstrate it??? " Where was that established ?? The question only says " Can you find the area of blue triangle ?? ", and doesn't mention to demonstrate theorems Everytime you apply "Pitágorean theorem" for example, you demonstrate it ????? Theorems have been already demonstated, centuries ago, we only apply those theorems !! Besides of that , I wrote you that modifying angles "α" and "β", we obtain the same rule..THAT ALSO is a demonstration !!! I write my comments just for fun, not to write a bibliography !!! If you want demonstration, just see video !!!, and not criticize to those who found the correct result !!
@@marioalb9726 We all understood that the height h of triangle ACQ is the radius of the circle inscribed in triangle ABC since Q is the intersection of the bisectors of ABC. But you do not prove that the square with side h has the same area as triangle PBR. You say area(PBR)=y*z/2=area(square with diagonal BQ)=h^2 So you say: y*z/2=h^2 quickly without proving. Your comment is 9 lines long to arrive at the result but is not a valid demonstration. @MathandEngineering makes a serious video of 14:36 which seems to be the best of the demonstrations. I propose other serious demonstrations which require a lot of lines.
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Good demonstration. Thanks too to @maisonville7656 for these interesting remarks.
Here another demonstration, a bit different from 48 hours ago before you deleted a similar video.
Like you at 8:22, Q is the incenter of the right-angled triangle ABC (Q center of the inscribed circle with radius=h)
Then :
h=(AB+BC-AC)/2
AB=AC*cos(2*a) and BC=AC*sin(2*a)
h=AC/2*(cos(2*a)+sin(2*a)-1)
PB=x/2 and BR=108/x then we must understand that : PB*BR=x/2*108/x=54
PB=BC*tan(b)=AC*sin(2*a)*tan(45°-a)
BR=AB*tan(a)=AC*cos(2*a)*tan(a)
PB*BR=AC^2*sin(2*a)*tan(45°-a)*cos(2*a)*tan(a)
We can notice that :
sin(2*a)*tan(a)=1-cos(2*a)
cos(2*a)*tan(45°-a)=1-sin(2*a)
PB*BR=AC^2*(1-cos(2*a))*(1-sin(2*a))
PB*BR=1/2*AC^2*(cos(2*a)+sin(2*a)-1)^2
PB*BR=1/2*(2*h)^2
PB*BR=2*h^2
h=sqrt(PB*BR/2)
h=sqrt(54/2)
h=3*sqrt(3)
A|ACQ|=1/2*AC*h
A|ACQ|=1/2*26*3*sqrt(3)
A|ACQ|=39*sqrt(3)
************************
Nota Bene :
We can demonstrate in other way, geometrically that : PB*BR=AC^2*(1-cos(2*a))*(1-sin(2*a))
Let's introduce 2 points :
S orthogonal projection of P on (AC)
T orthogonal projection of R on (AC)
Then, PBC and PSC are 2 identical right-angled triangle (same angles b, 90°-b, 90° and same hypothenuse PC)
Then, ABR and ATR are 2 identical right-angled triangle (same angles a, 90°-a, 90° and same hypothenuse AR)
Then : PB=PS ; BR=RT ; SC=BC ; AT=AB
TC=AC-AT=AC-AB=AC-AC*cos(2*a)=AC*(1-cos(2*a))
AS=AC-SC=AC-BC=AC-AC*sin(2*a)=AC*(1-sin(2*a))
AS*TC=AC^2*(1-cos(2*a))*(1-sin(2*a))
APS and RTC are 2 right-angled triangle.
PS=AS*tan(2*a) and RT=TC*tan(2*b)=TC/tan(2*a)
PS*RT=AS*tan(2*a)*TC/tan(2*a)
PS*RT=AS*TC
PB*BR=AS*TC
PB*BR=AC^2*(1-cos(2*a))*(1-sin(2*a))
Thank you sir for this amazing presentation
It's my pleasure sir
Q is the center of the in-circle. angle B being 90 degrees, BQ must be equal to sqr(2)*radius of in-circle. Figuring out that triangles BQR and BQP are similar was more work, but can easily be done. BR/BQ=BQ/BP gives 2*radius^2=54. The radium of the in-circle is the height of triangle AQC. Nice problem. Most of the work was to demonstrate BQR and BQP triangles are similar.
Yes sir, thanks. It's a good method you shared and personally I find it interesting. Thanks
Area of right triangle BPR:
A = ½ y.z = ½ .x/2 .108/x = 27 m²
Now, let's find the square adjacent to previous right triangle, with same area.
A = 27 m² = s² --> s =√27m
This square has opposite vertices B and Q, and its side is equal to height of blue shaded triangle:
h = s = √27 = 3√3 m
Area of blue shaded triangle:
A = ½b.h = ½ 26 3√3
A = 67,54 m² ( Solved √ )
[ There's a little mistake in video, y=x/2 m and not y=x/(2m) ]
That's weird.
You seem to demonstrate the exercise in a few lines, with too few calculations.
How do you prove that the square whose side is h, which therefore has [BQ] as diagonal, has the same area as the triangle BPR ?
@@matthieudutriaux
Thanks four your comment.
For any angle "α" and "β", while the triangle is a right triangle, then always BQ is the diagonal of a square, with sides aligned with legs of triangle. Because always location angle of BQ is 45° respect to right triangle legs.
And, of course, side "s" is equal to "h" easy to see, property of bisected triangles.
Regarding the area of the minor right triangle PBR, always Is equal to area of the mentioned square with diagonal BQ. You can find the demonstration in video, minute 13:11, where says " y.z= 2.h² ", what is the same I wrote in my comment.
This is a property of bisected right triangles, and we already know it in advance.
If you have any doubt, just modify the ratio y/z , or what is the same, modify angles "α" and "β". and you will verify that always this rule is fulfilled.
@@marioalb9726
I understand all the video of @MathandEngineering
Then, i understand the demonstration at 13:11 when "y*z=2*h^2" (little error : not y*x)
I have demonstrated myself (by 2 methods) that PB*BR=y*z=2*h^2.
But me and @MathandEngineering and others make the demonstration.
I am not agree with you when you say : "This is a property of bisected right triangles, and we already know it in advance."
All this exercise is a simple right-angled triangle.
Q is the center of bisectors.
Area |ACQ|=1/2*base*h=1/2*26*h
The difficulty of the exercise is to prove that y*z=2*h^2
I know we can change the ratio y/z and we will always have y*z=2*h^2
(but if we change the ratio y/z and if AC=26 then y*z=2*h^2 can be different from 54)
You may be right but you consider this relationship "y*z=2*h^2" as a mathematical theorem of "bisected right triangles" that we learn and know in advance.
The goal of the exercise is to demonstrate it.
If we already know it, it is true that the answer is faster.
@@matthieudutriaux
Why do you say "the goal of this exercise is to demonstrate it??? "
Where was that established ??
The question only says " Can you find the area of blue triangle ?? ", and doesn't mention to demonstrate theorems
Everytime you apply "Pitágorean theorem" for example, you demonstrate it ?????
Theorems have been already demonstated, centuries ago, we only apply those theorems !!
Besides of that , I wrote you that modifying angles "α" and "β", we obtain the same rule..THAT ALSO is a demonstration !!!
I write my comments just for fun, not to write a bibliography !!!
If you want demonstration, just see video !!!, and not criticize to those who found the correct result !!
@@marioalb9726
We all understood that the height h of triangle ACQ is the radius of the circle inscribed in triangle ABC since Q is the intersection of the bisectors of ABC.
But you do not prove that the square with side h has the same area as triangle PBR.
You say area(PBR)=y*z/2=area(square with diagonal BQ)=h^2
So you say: y*z/2=h^2 quickly without proving.
Your comment is 9 lines long to arrive at the result but is not a valid demonstration.
@MathandEngineering makes a serious video of 14:36 which seems to be the best of the demonstrations.
I propose other serious demonstrations which require a lot of lines.